The post What Is Entropy Information Theory In Calculus: Ultimate Guide appeared first on Calculus Help.
]]>Entropy is the concept of chaos and disorder. It affects everything, including information. Learn how entropy works in calculus.
You’ve likely heard of the concept of entropy, or at least the word. Entropy has different specific definitions depending on the context in which you’re talking about it, but the common thread among the definitions is that of uncertainty and disorder. Entropy in any given system tends to increase over time when not provided with external energy or change.
Entropy, simply put, is the measure of disorder in a system. If you look at an ice cube floating in a mug of hot tea, you can think of it as an ordered system. The tea and the ice cube are separate. Over time, however, the heat from the tea transfers into the ice, causing it to melt and mix with the tea and leaving a homogenous mixture. There is no way to restore it to its previous state.
Everything in existence is subject to entropy per the second law of thermodynamics. There are several other definitions of the word, but what we’re here to talk about is the entropy information theory in calculus.
Information is data about a system, an object, or anything else in existence. Information is what differentiates one thing from another at the quantum level. The entropy of information has more to do with probability than with thermodynamics: simply put, the greater the number of possible outcomes of a system, the less able you are to learn new information.
Following from this, it also means that a low-data system has a high amount of entropy. For example, a six-sided die when cast has an equal chance (unless you’re using weighted dice) of landing on any one of the six sides; the exact percentage of probability for any side landing face-up is 16.67 percent.
A 12-sided die would have a probability of 8.33 percent for landing face-up on any of the twelve sides. Its information entropy is even higher than that of the six-sided die.
As another example, if you have a regular coin with two sides, flipping it yields a 50 percent probability of landing on either heads or tails. The coin, because there is a higher probability for either outcome, has a lower level of entropy than does the die.
The specific entropy information theory in calculus we’re talking about refers to data systems that have random outcomes, or at least some element of randomness. The level of entropy present in a data set refers to the amount of information you can expect to learn at any given time.
As such, the only factor that truly influences the level of entropy in a data set is the number of possible outcomes or the specificness of the information. For example, if you have a group of numbers from 1 to 10, and your only criteria is that the number is even, you only get one bit of information. It is measured in bits just as data in computers is.
The equation used for entropy information theory in calculus runs as such:
H = -∑n_{i=1} P(x_{i})log_{b}P(x_{i})
H is the variable used for entropy. The summation (Greek letter sigma), is taken between 1 and the number of possible outcomes of a system. For a 6-sided die, n would equal 6. The next variable, P(x_{i}), represents the probability of a given event happening. In this example, let’s say it represents the likelihood of the die landing face-up on 3. Other factors being equal, you get 1/6.
Then, you multiply this by the logarithm base b of P(x), where b is whichever base you’re using for your purposes. Most often, you’ll use 2, 10, or Euler’s number e. For reference, e is approximately equal to 2.71828.
This, of course, represents only a single discrete variable with no other factors influencing it. If you have another factor that can influence the outcome, the formula changes. Now, it ends up being the following:
H = -∑_{i,j} P(x_{i}, y_{j})log_{b}P(x_{i}, y_{j})/p(y_{j})
In this case, you have to look at both x and y as variables and take their functions as dependent on one another, depending on how the problem is set up. You can also simply treat the two variables as two separate and independent events that occur. For example, f you were to flip two coins, you could treat x and y as each coin coming up heads or tails.
Alternately, you could assume that you had one coin coming up heads, which would trigger another coin flip whereas landing on tails would not. Then, you’d take y as the variable representing the second coin. Again, it varies depending on what you’re doing.
You can determine how much information is received from an event with a reduction of the formula to
-∑p_{i}logp_{i}.
You might notice that the log has no indicated base. This is because by default the base is 10. In computer science, when charting out probabilities, you might see base 2 used a lot because of binary systems. Base e is used in many scientific disciplines. Base 10, meanwhile, is used in chemistry and other sciences that aren’t quite as heavy on math.
Base 10, after all, is the basis of the decimal system and the one most people are used to working with.
The above formula represents the average amount of information you can expect to gain from an event per iteration. It assumes all factors are equal and there are no influencing conditions on the outcome. In other words, it is completely random within the available set of data.
Entropy information theory in calculus has several possible measurements, depending on what base is being used for the logarithm. Here are a few of the most common measurements:
On occasion, you may have to convert one measurement to another.
One question that inevitably arises when dealing with higher mathematics is why it should be studied. After all, many skills in advanced math lack real-world applications, at least to the uninitiated. However, it should be noted that information and probability theory have several applications in computer science, such as file compression and text prediction.
For example, by using the entropy theory of information, you can help to code predictive text. The English language, for instance, has many different rules about what letters can occur in a sequence. If you see the letter ‘q’, you know that in all likelihood it’s followed the letter ‘u’. If you have two vowels, they will likely not be followed by a third, and certainly not by a fourth.
This is just one example. Another possible use in computer science is data compression as in ZIP files or image files like JPEGs. Information theory has also encompassed various scientific disciplines from statistics to physics. Even the inner workings of black holes, such as the presence of Hawking radiation, rely on information theory.
Even the act of learning higher math and abstract concepts can be beneficial, even without the knowledge. The act of learning forces the brain to process information in new ways, creating stronger connections between neurons and delaying the onset of cognitive decline. The more you learn, the more exercise your brain gets.
To get the most out of and understand information theory, you need a solid grasp on a few other subjects. These are all unsurprisingly math-related. Calculus and statistics are the two main subjects you need to study before starting to work on information theory because you’ll need to know how to take integrals and derivatives from calculus.
Statistics, meanwhile, gives you a solid understanding of probability theory and the likelihood of events occurring. You’ll also be introduced to a few of the more esoteric variables like lambda, or some of the symbols like sigma for summation if you haven’t seen them before.
You’ll also need to have a solid grasp of algebra, such as knowing how to manipulate equations and variables. Although you should get plenty of practice doing this in the course of algebra, it’s a good idea to understand some other concepts such as how to take a logarithm. As you can see, logarithms play a vital part in several of the entropy equations.
Learning the entropy information theory in calculus is a good way to understand how probability works and how many of the data systems you encounter produce various amounts of information. If you have a background in thermodynamic studies, it can make it easier to understand the concept of entropy.
Entropy in information theory is slightly different than it is in other branches of science, but the basic idea is the same.
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]]>The post Problem 3: Position and Velocity appeared first on Calculus Help.
]]>An object is dropped from the second-highest floor of the Sears Tower, 1542 feet off of the ground.
(a) Construct the position and velocity equations for the object in terms of t, where t represents the number of seconds that have elapsed since the object was released.
(b) Calculate the average velocity of the object over the interval t = 2 and t = 3 seconds.
(c) Compute the velocity of the object 1, 2, and 3 seconds after it is released.
(d) How many seconds does it take the object to hit the ground? Report your answer accurate to the thousandths place.
(e) If the object were to hit a six-foot-tall man squarely on the top of the head as he (unluckily) passed beneath, how fast would the object be moving at the moment of impact? Report your answer accurate to the thousandths place.
(a) The position function for a projectile is s(t) = –16t2 + v0t + h0, where v0 represents the initial velocity of the object (in this case 0) and h0 represents the initial height of the object (in this case 1,542 feet). Note that this position equation represents the height in feet of the object t seconds after it is released. Thus, the position equation is s(t) = –16t2 + 1,542. The vecocity equation v(t) is the derivative of the position equation: v(t) = –32t.
(b) Average velocity is the slope of the secant line, rather than the slope of the tangent line. Finding average velocity is easy. Plug t = 2 and t = 3 into the position equation to calculate the height of the object at the boundaries of the indicated interval to generate two ordered pair: (2, 1478) and (3, 1398). Apply the slope formula from basic algebra to calculate the slope of the line passing through those points.
(c) Substitute t = 1, 2, and 3 into v(t).
(d) The object hits the ground when its position is s(t) = 0. Set the position equation equal to zero and solve for t.
(e) The problem asks you to calculate the velocity of the object when it is exactly six feet off of the ground, when s(t) = 6. Apply the same technique you completed in part (d), but instead of calculating the time t when the object’s position is 0, calculate the time t when its position is 6.
Now calculate the velocity of the object at that time: v(9.79795897113) = –32(9.79795897113) = –313.535 ft/sec.
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]]>The post Ultimate Guide On How To Calculate The Derivative Of Arccos appeared first on Calculus Help.
]]>The derivative of arccos in trigonometry is an inverse function, and you can use numbers or symbols to find out the answer to a problem. It uses a simple formula that applies cos to each side of the equation. Some people find using a drawing of a triangle helps them figure out the solutions easier than using equations.
The derivative of Arccos is used in trigonometry. It’s an inverse function, and you can manipulate it with numbers or symbols. There are several terms you’ll need to know when working with Arccos, including radian.
Arccos means arccosine. You may also work with arcsin, or arcsine when working with trigonometry problems.
Results for the derivative of Arccos are written in radians. Solving a problem involving Arccos first requires knowledge of radians.
Radians are used to measure angles. The angle of a single radian subtends or creates an angle. At a certain point, extremities present straight lines that join at that particular point. When two rays pass through the arc endpoints, the angle is subtended.
The radian has an angle of a single radian subtended from a unit circle’s center. A radian creates an arc length of 1. Therefore, we can determine that a full angle measures 2pi radians. The angle contains 360 degrees for every 2pi radians, which is equivalent to 57.29577951 degrees per radian or 180 degrees pi.
A right angle measures pi/2 radians. A straight angle measures pi radians. Radians let you write integrals and derivative easily, like d/(dx)six equals cosx when x is measured by radians.
The three branches of mathematics are algebra, arithmetic, and geometry. Geometry is the study of sizes, properties, and shapes of spaces of three- dimensional and two-dimensional objects. Euclid, the Father of Geometry, is often mentioned in conjunction with theorems and postulates about the field.
Geometry is a combination of the Greek terms Geo (earth) and metron (measure). You will encounter solid, spherical and plane geometry in your studies.
Spherical geometry is the study of three-dimensional objects such as spherical polygons and spherical triangles. Plane geometry features lessons about two-dimensional objects such as points, curves, lines, and planes, including polygons, circles, and triangles. Solid geometry focuses on spheres, prisms, cubes, pyramids, and other three dimensional objects.
Euclidean Geometry is the study of flat surfaces, and it’s just one of the branches of geometry. Riemannian geometry, the study of curved surfaces, is another major branch of geometry.
Trigonometry is a branch of geometry. The field emerged in about 150 B.C. when a mathematician named Hipparchus used sine to make a trig table.
Trigonometry deals with triangles, their lengths, and their angles. You can also use trigonometry to study waves and oscillations. You will study the way the side lengths of right triangles relate to each other. The basic relationships between triangles and their sides are referred to as Sine, Tangent and Cosine.
If you have a right angle triangle, the longest base is referred to as the hypotenuse. The side in front of an angle is its opposite side, and the adjacent side is the side behind that angle. The relationship for a right angle triangle then is sin A, or the opposite of the hypotenuse, cos A, the adjacent side of the hypotenuse, and tan A the opposite/adjacent side.
The secondary relationships in trigonometry are:
These measurements are the respective dimensions of Sine, Cosine, and Tangent.
Spherical trigonometry that deals with triangles in 3D scenarios, and it’s used in navigation and astronomy.
Trigonometry helped ancient sailors navigate the seas. Today, trigonometry has many uses beyond math class.
Oceanographers use trigonometry to figure out the height of ocean tides or measure sea animals. Trigonometry can be used to build ships and help them navigate. This mathematical field also helps criminologists determine the trajectory of bullets and other projectiles.
Archeologists divide their excavation sites up for work using trigonometry. Flight engineers use trigonometry to determine the best flight course for a plane. Trigonometry helps fill in the “third side” of a flight’s wind, speed, and direction equation to make sure the plane goes in the right direction.
In trigonometry, the Arccos means cos to the negative first power (z) or the arc cosine of the complex number represented by (z). You will consider that this problem involves an inverse trigonometry function and write down two identities:
Cos (cos to the negative 1^{st }power(x)) equals x, then cos to the negative 1^{st} (power (cos(x)) equals x.
You need to decide which formula to use. You should use the first formula, due to the chain rule. The chain rule allows you to find the derivative for the term cos−1
(x) more easily than using the first formula. Simply the formula to cos (y) equals x. Now you have an implicit formula to solve the problem and determine y.
Use implicit differentiation to find out the answer for y’.
(cos(y))0 = (x)followed by y’ (− sin(y)) = 1 and y’ equals negative 1 over sin (y), then y’ equals negative 1 over sin(cos to the negative 1^{st} power(x)).
The formula written above is an excellent tool, but there is an even better way to write it using geometry instead of algebra. Cos has an angle and a number between negative 1 and 1, cos negative one will also give angle and have a number between negative 1 and 1.
Now define θ equals cos minus 1(x)to show the angle cos negative 1. Therefore, x equals cos (θ), according to the inverse function. You can also use cos with both sides of the function.
Drawing a triangle can also help you find the answer to this problem. Draw a triangle with points ABC and θ will be the angle less than C. Cos (θ) equals ACBC in geometry, and cos (θ) also equals x.
Consider that BC equals 1 and AC equals x. You could choose any identity in the above values, but the most obvious ones make the math easier.
In the triangle example, sin(θ) = AB over BC. Use the Pythagorean Theorem to calculate the answer for sin(cos−1(x)), which is equal to sin(θ).
Here’s the solution per the Pythagorean Theorem:
BC squared equals AB squared plus AC squared, followed by AB squared equals BC squared minus AC squared. Therefore, AB squared equals one minus x squared, and AB equals the square root of 1 minus x squared.
The equation calculated above answers the question correctly. Sin(cos negative 1 (x)) equals sin(θ) equals AB over BC equals AB equals the square root of 1 minus x squared. Complete the answer with
This problem demonstrates how to determine the derivative of Arccos x and Arcsin x.
The equation you use is d over dx (arcsin x plus arccos x) equals zero. Note that arcsin x plus arccos x equals pi over 2. You can explain this equation with the following calculations:
If arcsin x equals zero, then x equals sinθ equals cos( pi over 2 minus θ), then arccos x equals pi over 2 minus θ equals pi over 2 minus arcsin x; therefore arcsin x plus arccos x equals pi over 2.
The derivative you find for arccos x will be the negative of arcsin x’s derivative. This derivative stature holds for the inverse of each cofunction pair.
The same formulas apply to similar trigonometry problems. Arccot x’s derivative is the negative of arctan x’s derivative. Arcsec’s derivative is the negative of the derivative of arcsecs x.
The variable y equals arcsec x, represent tan y equals plus-minus the square root of x to the second power minus one.
Begin solving the problem by using y equals arcsec x, which shows sec y equals x. Now use Pythagorean identity b to figure the next step.
Tan y equals plus minus the square root of sec squared y minus 1 equals plus minus the square root of x squared minus 1. Y’s derivative equals arcsin x.
The equation then is d over dx arcsin x equals one over the square root of one minus x squared. Prove this by looking at y equals arcsin x, which stands for sin y equals x. Figure the derivative of x with the following equation:
Cos y followed by dy over dx equal 1, then dy over dx equals 1 over cos y’, then dy over dx equals 1 over the square root of 1 minus x squared ‘.
An alternate theorem for this would be d arcsin x over dx equals 1 over din sin y over dy equals 1 over cos y equals 1 over the square root of 1 minus x squared.
The arccosine of x is the representation of the inverse cosine function of the variable x when negative 1 is less than x and x is less than 1. If the cosine of y equals x, the arccosine of the variable x is equal to x’s inverse cosine function, which equals y. The equation for this function reads arccos x equals cos to the inverse cosine x equals y.
The equation is written arccos 1 equals cos, followed by the negative 1 symbol, which stands for inverse cosine function in this problem.
Therefore, arccos 1 equals cos negative 1 (inverse function), and 1 equals zero, and Rad equals zero degrees.
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]]>The post Finding Limits In Calculus – Follow These Steps appeared first on Calculus Help.
]]>If you’ve been asked to find a limit by your teacher, there are a range of different methods that you can use. It’s much simpler than it sounds and at the end of this guide, we have a nifty strategy that you can follow so that you’ll always know which method to use and when.
Finding limits isn’t easy, and a lot of people struggle with it. If this is you, don’t worry, by the end of this guide, we’ll have you finding limits in a few minutes at most. Read carefully and try to follow along, but don’t be afraid to start from the beginning to perfect your technique and to memorize each of the strategies properly.
Limits are incredibly essential, and without them, we would be unable to do more advanced forms of calculus. A limit is the limit of a function f(x) as x approach c but never reaches it. Remember, x can approach c from either side. Picture a graph; it can come from either side of the axis.
Limits allow us to find out how a function will behave even if it doesn’t exist at a specific value of x. The result is that finding limits will allow you to derive the angle of a slope at a given point, even if you don’t have a specific value of x for every point along the line. Without knowing how to find limits, we would have little information about the gradient between points.
If we take the function f(x) = x – 1 / x – 1 and then imagine that x could be any number.
We know that if x = 1 the function would look like this:
f(1) = 1 – 1 / 1 – 1 which would equal 0 / 0.
Image by Edmund Fung from Pixabay
The result is that when x = 1, the function itself is undefined because the fraction 0 / 0 is undefined. On a graph, this would look like a straight line across, parallel to the x-axis, but there would be a gap where x = 1 because it’s just not defined. But what if we wanted to know what the function was when x = 1?
Well, we can’t do it. But what we can do is to get as close as possible to x = 1 so that we can know approximately what the value of the function is at that point. This idea is necessarily a limit. It’s the idea of being able to get as close as possible to an undefined point so that we can approximate it with a high level of accuracy.
The first technique that we’ll look at is plugging x into the function to see the limit. In an ideal world, this would work all of the time. Therefore, we always start with this technique because it’s the simplest and allows us to get more information about what to do next. The idea is that you make x equal to the number it ’s approaching.
So, if we are trying to find the limit as we approach 2, we make x = 2 and then run the function.
When you do this, you’ll get one of three results:
f(a) = b / 0 where b is not zero.
f(a) = b where b is a real number.
f(a) = 0 / 0.
In the first circumstance, you’ve probably found an asymptote. An asymptote is when a line continually approaches a given value, but it will never reach it at any finite point.
In the second situation, you have probably found the correct limit through the substitution method.
Finally, in most complicated questions you will end up with a situation where the function is undefined, and therefore you’ll need to try other techniques. If this is the case, we will need to rearrange the function so that we can consider the limit in an identical but differently arranged form using one of the following three techniques.
Factoring is a great method to try and is often one of the easiest to learn because it relies on skills that you’ve already practiced. If you’ve already tried to plug in a number have ended up with 0 / 0, you need to start factoring.
Often you’ll see that either the numerator or the denominator is more ‘friendly’ to factoring. Usually, x with the highest power is the best place to start. Let’s consider the following equation:
x^2 – 6x + 8 / x – 4 where x is approaching 4.
In this example, the numerator is the only place for you to factor. It’s also obvious because of the x^2 which can factor. In this case, we can factor to:
(x – 4)(x – 2) / (x – 4)
As you can see, we can then cancel the two matching x – 4 on both the top and the bottom. Pretty simple, right? It won’t also be this easy, but if you continue to factor you can often find places to simplify the expression.
This simplification leaves us with:
f(x) = x – 2 where x is approaching 4.
If we try to substitute 4 into the equation now, you’ll find the f(x) = 2. See, by factoring you’ve shown that the equivalent function has a specific value and that value is 2 when x is approaching 4.
If you were to create a graph of this function, you would still see a gap where x = 4 because the original equation is still undefined. However, you’ll know that when approaching 4, the function equals 2.
After factoring, you might find that there is no way for you to cancel and simplify. In this case, you should try another method to ensure that there is no limit of the function at the specific value of x.
The third technique requires you to rationalize the numerator so that you can try substitution again. You’ll know if you should rationalize the numerator because you’ll see a square root on the top and a polynomial expression on the bottom. Let’s look at the following example:
f(x) = sqr(x-4) – 3 / x – 13 as the function approaches 13.
We know that substitution fails when you get 0 in the denominator, and therefore substitution would fail in this example. Factoring would also fail because there is no polynomial to factor in this example.
However, if you were to multiply the numerator and denominator by the conjugate of the top (numerator), then you’ll be able to cancel and find the limit.
The conjugate of the numerator is: sqr(x – 4) + 3 and therefore we can multiply through to get:
(sqr(x – 4) – 3)(sqr(x – 4) + 3) / (x – 13)(sqr(x – 4) + 3)
We can then FOIL the numerator to get the following:
(x – 4) + 3sqr(x – 4) – 3sqr(x – 4) – 9
When simplified the above expression will become x – 13 because the middle terms cancel and then you can combine like terms.
If we go back to the full equation you can now see that we have:
(x – 13) / (x – 13)(sqr(x – 4) + 3)
The terms cancel, and we have:
1 / (x – 13)(sqr(x – 4) + 3)
From there, we can plug in 13 into the function because we have all of the unknowns on one side of the fraction. The result is that the limit is ⅙.
So far we’ve only looked at situations which don’t include any trigonometry. These require unique methods like factoring and conjugates to ensure that you can simplify and be able to easily plug in a number for x to find the limit. We want to do the same with this equation, but it contains trigonometry which complicates things a little.
For you to solve these equations, it’s vital that you know all of the trig functions so that you can rewrite equations and more effectively address them.
The most common are as follows:
Cos (x) = 1 / Sin (x)
Sec (x) = 1 / Cos (x)
Cot (x) = 1 / Tan (x)
It’s highly likely that you’ll also need to know the double angle identities in order to simplify more complex functions.
Sin (2a) = 2Sin(a)Cos(a)
Cos (2a) = Cos^2(a) – Sin^2(a)
Tan (2a) = 2Tan(a) / 1 – Tan^2(a)
This equations might seem confusing, but they are actually very simple. They are each used for different purposes, but when finding limits we only need to know them for rewriting equations.
Let’s look at the following example:
Sin (x) / Sin (2x) when x is approaching 0
We can use the double angle identities formula to simplify to:
Sin (x) / 2Sin(x)Cos(x)
From there, the Sin(x) can cancel and we are left with:
1 / 2Cos(x)
If we plug in 0 as x, we will get ½ because cos(0) = 1 and therefore you have 1 / 2*1 which is ½.
Now that we’ve covered all of the tactics that you can use to find limits let’s discuss which you should use and when. There is a straightforward rule. You should always do a direct substitution first.
If you get f(a) = b / 0 then you have an asymptote.
If you get f(a) = b then you have a limit.
If you get f(a) = 0 / 0 then you should try factoring, rationalizing the numerator or trig identities depending on which seems most likely to work.
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]]>The post How To Solve A Logarithmic Equation In Calculus appeared first on Calculus Help.
]]>Image source: Pixabay.com
A logarithmic equation needs to be rewritten as an exponential equation for you to find the variable. This common calculus problem contains constants and expressions, and you’ll find logarithmic shortened to “log” in some written problems.
Condense a problem with more than one logarithm by turning it into one equation. A logarithm is constructed in a way that allows it to change a math function into another math function to solve a problem. Multiplication turns into addition and division becomes subtraction. The function changes let you find the variable value.
You’ll work with some well-known calculus rules with logs, such as:
The Power Rule
The Product Rule
The Quotient Rule
You will work with many components to solve logarithmic equations. Here are a few of them, with some brief information about the function of each one.
An exponent is a small raised number next to a larger numeral in an equation. The small number shows how many times the larger number is multiplied by itself. The small raised number 2 shows that the larger number is multiplied by itself twice. A number multiplied by itself twice can be referred to as 4 to the second power or four squared.
A variable is a symbol for an unknown number. Linear equations may have many values that can be used in place of the variable. Most variables, however, are solved with a single value.
The variable F represents the graph of a function. The graph of a function represents all points in f(x). You can also refer to the graph of a function as the graph of an equation.
When you look at the equation of a straight line (such as y equals mx plus b) the y-intercept is the location where the line goes through the vertical y-axis. In y equals mx plus b, b is the y-intercept, the slope is m, and the variable m is multiplied on the variable x.
An x-intercept appears at the spot where the graph crosses the x-axis. The x-intercept is also the point on the graph that shows the variable x as zero.
A problem with one logarithm on each side of the equation that has the same base lets you use arguments that are the same. The expressions M and N are the arguments in the following description:
Log_{b }M equals Log_{b }N leads to M equals N
A problem with a logarithm on one side of the equation you can use an exponential equation to find the answer.
Log_{b} M equals N leads to M equals b^{N}
Solve this problem:
Log_{3 }(x) plus Log_{3 }(x -2) equals Log_{3} (x plus 10)
Condense the log arguments on the left side into one logarithm with the Product Rule. You need one log expression on both sides of the equation. X will have a power of two, so you’ll need to solve a quadratic equation.
You get Log_{3 }[(x) (x minus 2)] equals Log_{3 }(x plus 10). Now condense (x) (x-2) = x squared minus 2x. Then you get Log_{3 }(x squared minus 2x) equals Log_{3 }(x plus 10)
Get rid of the logs and set the arguments inside the parenthesis to match each other.
X squared minus 2x equals x plus 10
Now use the factoring method to finish the quadratic equation. Move all information to one side and make the opposite side contain a zero value.
X squared minus 3x minus 10 equals zero
(x minus 5) (x plus 2) equals zero
Now turn each factor to zero and solve x. X minus 5 equals zero means that x equals 5. X plus 2 equals zero means that x equals negative 2. X equals 5 and X equals negative 2 are the answers we may use. Now check the answers to see if they are correct.
Place the answers back in the first logarithmic equation to verify their validity.
For x equals 5, Log_{3 }(5) plus Log_{3} (5-2) equals Log_{3 }(5 plus 10)
Log_{3 }(5) plus Log_{3 }(3) equals Log_{3 }(15)
This is the correct answer. X equals negative 2 gives us a few negative numbers inside the parenthesis, and a log of zero and negative numbers in the equation make negative 2 the wrong answer.
Solve the following problem: ½ log (X to the fourth power) minus log (2x minus 1) equals log (x squared) plus log (2).
Log without a written base has a base of 10. Base 10 is the common logarithm. Compress both sides of the equation into one log. You’ll see the Quotient Rule applied on the left side ( a difference of logs) and the Product Rule (the sum of logs) on the right side.
Pay attention to the ½ coefficient on the left side. You’ll need to use the Power Rule and bring the coefficient ½ up in reverse order.
Log (base), M to the k power equals K times log (base) M, then ½ log (X to the fourth power) minus log (2x minus 1) equals log (x squared) plus log (2)
Now use the ½ as an exponent on the left. Log (x to the fourth power) to the one/half power minus log (2x minus 1) equals log (x to the second power) plus log (2).
Now simplify the exponent to log (x squared) minus log (2x minus 1) equals log (x squared) plus log (2) Now condense log using the Product Rule on the right and the Quotient Rule on the left.
Log ( x squared over 2x minus one) equals log (2x to the second power)
It’s all right to show that if we have the same base in our equations (base 10), we can show that they are equal to each other. Now drop the logs and put arguments inside their parentheses.
X squared over 2x minus 1 equals 2x squared
Use the Cross Product to solve the Rational Equation. Factor out to get the brief, final answer after moving all terms to one side of the equation. Make each factor equal to zero and then solve x. X equal ¾ is one possible answer, x equals zero is the other.
Check your possible answers. X equals zero bring an undefined zero logarithm into the equation, which is wrong. X equals 3.4 is the only solution.
Students who consistently get excellent grades in calculus or geometry do so because they study daily and show a steady interest in their classwork and homework. Establish a study routine by choosing a quiet place where you can read and practice solving problems at the same time each day (or at least a few times a week).
Bookmark math websites and videos containing videos that will help you understand the formulas and concepts that give you trouble, including any logarithmic equation. Along with your class notes and practice problems in your textbook, you’ll be equipped with everything you need to study more efficiently.
Contact other students in your class, tutors from the math department, or independent math tutors to help you if you’re unable to master a particular formula on your own. Even the best students need help from another student or a tutor from time to time. You can find help online from professional tutors who are available 24/7 to help answer your questions. To ask a question click here.
Don’t expect calculus or any math class to be difficult or easy; work on the assignments without worrying about your grades or the outcome. Study for tests up to a week in advance, preferably with classmates or other math students. Trade tips and discuss different solutions and approaches to problems.
No one fails calculus because they lack the skills or mental capacity to perform exercises properly. People fail because they are unwilling or unable to do the required work. Teachers and calculus experts suggest you study six or seven hours on weekends and a few hours each weeknight to get the best grades possible.
Anyone who only has five to ten hours a week to study and prepare for class or tests should delay calculus classes until they are ready to study more often. Calculus courses are fast-paced, and if you get behind it will be hard to get caught up on the lessons you missed or didn’t understand.
Work as hard as you can in the first month of class. Define your strengths and weaknesses, and enlist tutors or a study group if you need them. Don’t get behind and think you can catch up quickly without help. You may end up dropping the class if you don’t take charge of your studies.
Don’t miss classes. Calculus isn’t like history or English, where catching up is hard, but not impossible, by yourself. If you miss one class, you should be able to catch up if you have a passable grasp of previous classes.
Anyone with a poor understanding of previous formulas and problem-solving methods will find that they are completely lost after missing a class or two. As soon as you feel confused, let your teacher, tutor, or study group know and ask for help. If you want flexibility with help, then you can use online tutors.
Click here to get tutor help.
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]]>The post What Is the Tangent Line Equation? appeared first on Calculus Help.
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In calculus you will inevitably come across a tangent line equation. What exactly is this equation? This article will explain everything you need to know about it.
In calculus, you learn that the slope of a curve is constantly changing when you move along a graph. This is the way it differentiates from a straight line. You can describe each point on a graph with a slope.
A tangent line is just a straight line with a slope that traverses right from that same and precise point on a graph. When we want to find the equation for the tangent, we need to deduce how to take the derivative of the source equation we are working with.
When looking for the equation of a tangent line, you will need both a point and a slope. You will be able to identify the slope of the tangent line by deducing the value of the derivative at the place of tangency. This is where both line and point meet.
In regards to the related pursuit of the equation of the normal, the “normal” line is defined as a line which is perpendicular to the tangent. This line will be passing through the point of tangency. Now that we have briefly gone through what a tangent line equation is, we will take a look at the essential terms and formulas which you will need to be familiar with to find the tangent equation.
Before we get to how to find the tangent line equation, we will go over the basic terms you will need to know. By having a clear understanding of these terms, you will be able to come to the correct answer in your search for the equation.
With the key terms and formulas clearly understood, you are now ready to find the equation of the tangent line. You should retrace your steps and make sure you applied the formulas correctly. Otherwise, you will get a result which deviates from the correctly attributed equation.
When we are ready to find the equation of the tangent line, we have to go through a few steps. If you take all these steps consecutively, you will find the result you are looking for.
There is more than one way to find the tangent line equation, which means that one method may prove easier for you than another. We will go over the multiple ways to find the equation. The following is the first method.
While you can be brave and forgo using a graph to illustrate the tangent line, it will make your life easier to graph it so you can see it. This is because it makes it easier to follow along and identify if everything is done correctly on the path to finding the equation. You will want to draw the function on graph paper, with the tangent line going through a set point.
What you will want to do next is take the first derivative (f’x), which represents the slope of the tangent line somewhere, anywhere, on f(x), as long as it is on a point.
Take the point you are using to find the equation and find what its x-coordinate is. When you input this coordinate into f'(x), you will get the slope of the tangent line.
What you need to do now is convert the equation of the tangent line into point-slope form. The conversion would look like this: y – y1 = m(x – x1). In this equation, m represents the slope whereas x1, y1 is a point on your line. Congratulations! You have found the tangent line equation.
While you can be fairly certain that you have found the equation for the tangent line, you should still confirm you got the correct output. It helps to have a graphing calculator for this to make it easier for you, although you can use paper as well. You will graph the initial function, as well as the tangent line. If confirming manually, look at the graph you made earlier and see whether there are any mistakes.
There are a few other methods worth going over because they relate to the tangent line equation. Knowing these will help you find the extreme points on the graph, the equation of the normal, and both the vertical and horizontal lines.
With this method, the first step you will take is locating where the extreme points are on the graph. These are the maximum and minimum points, given that one is higher than any other points, whereas another is lower than any points. Remember that a tangent line will always have a slope of zero at the maximum and minimum points.
A caveat to note is that just having a slope of 0 does not completely ensure the extreme points are the correct ones. To be confident that you found the extreme points, you should take the following steps:
The “normal” to a curve at a specific point will go through that point. However, its slope is perpendicular to the tangent. When you want to find the equation of the normal, you will have to do the following:
To find out where a function has either a horizontal or vertical tangent, we will have to go through a few steps.
There are two things to stay mindful of when looking for vertical and horizontal tangent lines. In the case of horizontal tangents, you will want to make sure that the denominator is not zero at either the x or y points. In the case of vertical tangents, you will want to make sure that the numerator is not zero at either the x or y points.
Congratulations on finding the equation of the tangent line! You can now be confident that you have the methodology to find the equation of a tangent. It may seem like a complex process, but it’s simple enough once you practice it a few times. The key is to understand the key terms and formulas. Having a graph as the visual representation of the slope and tangent line makes the process easier as well.
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]]>The post Mean Value Theorem for Integrals: What is It? appeared first on Calculus Help.
]]>Image from Pixabay
The mean value theorem for integrals is a crucial concept in Calculus, with many real-world applications that many of us use regularly. If you are calculating the average speed or length of something, then you might find the mean value theorem invaluable to your calculations.
Ultimately, the real value of the mean value theorem lies in its ability to prove that something happened without actually seeing it. Whether it’s a speeding vehicle or tracking the flight of a particle in space, the mean value theorem provides answers for the hard-to-track movement of objects.
Based on the first fundamental theorem of calculus, the mean value theorem begins with the average rate of change between two points. Between those two points, it states that there is at least one point between the endpoints whose tangent is parallel to the secant of the endpoints.
A Frenchman named Cauchy proved the modern form of the theorem. One of the most prolific mathematicians of his time, Cauchy proved the mean value theorem as well as many other related theorems, one of which bears his name.
It is also possible for a function to have more than one tangent that is parallel to the secant. The derivative, or slope, of each tangent line, is always parallel to the secant in the mean value theorem.
As an addition to the mean value theorem for integers, there is the mean value theorem for integrals. This theorem allows you to find the average value of the function on at least one point for a continuous function.
Stipulations for this theorem are that it is continuous and differentiable. That means that the line acts as a traditional function, without any odd stops, gaps, drop-offs, or any other non-continuous feature.
It also must be differentiable, which means you can find the slope of a point on the function. For cube roots or the absolute value of x, you cannot find a derivative because they are either undefined or not tangential to the average rate of change.
Floor and ceiling functions also do not have derivatives because they are not continuous functions. Thus the mean value theorem of integers does not apply to them.
Like many other theorems and proofs in calculus, the mean value theorem’s value depends on its use in certain situations. Since this theorem is a regular, continuous function, then it can theoretically be of use in a variety of situations. Any instance of a moving object would technically be a constant function situation.
Real-world applications for the mean value theorem are endless, and you’ve probably encountered them either directly or indirectly at some point in your life.
One of the classic examples is that of a couple of police officers tracking your vehicle’s movement at two different points. You are then issued a ticket based on the amount of distance you covered versus the time it took you to complete that distance.
You were not speeding at either point at which the officer clocked your speed. But, they can still use the mean value theorem to prove you did speed at least once between the two officers.
More specifically, consider modern-day toll roads. These roads have cameras that track your license plate, instantaneously clocking your time spent on the road and where and when you exited and entered. Law enforcement could quickly begin to crack down on speeding drivers on these roads, by merely finding the average rate of change between the two points. Drivers could then blame the mean value theorem of integers as the reason for their ticket.
Further use occurs in sports, such as racing. When investigating the speeds of various racing objects, such as horses or race cars, technicians and trainers need to know the performance of horses or race cars at specific points during the race.
Using data obtained throughout the race, individuals can determine how their horse or car was performing at certain times. For horses, this can mean altering training patterns or other variables to improve performance related to results. Race car drivers can use the data to tune equipment in various ways to better utilize the car’s speed.
The critical part of the theorem is that it can prove specific numbers. It can determine the velocity of a speeding car without direct visual evidence, or the growth, length, and myriad other instances where an object or thing changes over time.
In the real-world, a continuous function could be the rate of growth of bacteria in a culture, where the number of bacteria is a function of time. You could divide the difference in the number of bacteria by time to find out how fast they multiplied.
Applying the mean value theorem to the above situation would allow you to find the exact time where the bacteria multiplied at the same rate as the average speed. This might be useful to researchers in various ways, to determine the characteristics of certain bacteria.
Another more practical situation would be to determine the average speed of a thrown baseball. The distance of the ball thrown is a function of time. Dividing the difference in the length by the time it took for the ball to get from point a to point b would tell you how fast the ball goes.
When the mean value theorem is applied, a coach could analyze at which point the ball achieved the average speed. If the speed was faster before or after the tangential point, then the coach could alter the mechanics or delivery of the player’s throw. This would make for more optimal speed with the throw reaches the batter.
Finally, let’s find the average speed of the vehicle and then at which point during the drive, the car reached a speed equal to the average rate.
When using the mean value theorem in practical applications like vehicle speed, it is essential to note that the average rate of change is just that – an average. If your vehicle speed is 50 mph, then at some point during your drive you drove over and under 50 mph. Of course, you would hit that speed at least twice at a minimum.
Another exciting application of the mean value theorem is its use in determining the area. When the point at which the tangent line occurs is understood, draw a line from the new point parallel to the x-axis. This line is the top of your rectangle. The bottom is the x-axis. The left side is the y-axis, and the right is the endpoint of your continuous function.
Once this is complete, the area of your rectangle will be the same as the area beneath the curve of your function. One practical application of this instance is determining the exact height of a liquid in a container. If the liquid is suspended or not at rest, then calculating the mean value theorem of integers for the endpoints of the liquid will help you to determine the resting volume.
Determining amounts of liquid or the properties of a substance are just a few of the many applications of the mean value theorem. All fields of science use this theorem, and merely finding the volume of a liquid at rest is just scratching the surface. As sport becomes more science-based, the value of this theorem will only continue to increase.
While a fundamental calculus theorem may not change your life, it can make your life a tiny bit more manageable. Understanding the movement of an object and the properties within that movement can help you make a variety of educated conclusions.
When working in scientific fields such as physics or biology, the use of the theorem can aid in the research of particles or microscopic organisms. In sports, you can use the theorem to develop a better understanding of fast-moving objects. On the highway, the police can issue more speeding tickets. The mean value theorem of calculus is an invaluable tool for all types of people.
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]]>The post How To Calculate Hyperbolic Derivatives appeared first on Calculus Help.
]]>A hyperbolic derivative is a derivate of one of the hyperbolic functions, which are functions that utilize the exponential function (ex) to simplify otherwise complex calculations. Common uses for hyperbolic functions include representing the length of arcs such as those formed by the cables of a suspension bridge, or the shape of the Gateway Arch in St. Louis, Missouri.
As derivatives calculate the rate of change of a function (such as deriving a function for a car’s acceleration given a function of its velocity over a variable of time), calculating a hyperbolic derivative means calculating the rate of change of a hyperbolic function. This definition is especially useful for understanding hyperbolic integrals, which are helpful in arc length calculations and other applications of mathematics and engineering.
If some of these terms are unfamiliar (or outright intimidating), don’t worry! We’ll give a quick review of some of the basics before diving into the actual derivations.
Hyperbolic functions are closely related to trigonometry, as the functions that they describe relate to a hyperbola in the same way that trigonometric functions relate to a unit circle. But what exactly is this relationship, or even the “unit circle” for that matter? Is there a “unit hyperbola?”
Trigonometry helps describe the “unit circle” represented by function x2+y2=1. On a two-dimensional graph, this circle has its center at the origin (point (0,0)) with a radius of one unit. Trigonometry is the study of the dimensions within this circle, where special functions provide the height and width of the circle’s radius at a certain angle. The “certain angle” part is especially important.
This concept can be challenging to visualize internally, but consider an easy example: Suppose that the unit circle’s “certain angle” starts on the x-axis at zero degrees. In other words, if we position a line within the unit circle directly on the x-axis, it will be at zero degrees relative to the rest of the circle. Rotating it about the origin towards the positive y-axis will increase this angle in the positive direction while turning it towards the negative y-axis will decrease this angle in the negative direction.
If we leave our imaginary line on the x-axis, what are the “components” of its height and width? Well, at zero degrees (or the x-axis), its width is its length. So, if our imaginary line has the same radius as the unit circle (one unit), then we could say that the “width” component of our circle is also one unit.
However, what about our height component? Where do we start measuring? We measure the height component of our line from the x-axis, just as we measured its width from the y-axis. So, since our line is lying directly on the x-axis, its height component must be zero.
As we rotate our imaginary line inside the unit circle, the height and width components of our imaginary line will change with the angle of rotation. Laying the line directly on the positive y-axis, for example, will have height and width components of one and zero units, respectively. But what about points in between?
Thankfully, the height and width components of our line are conveniently described using basic trigonometric functions. Here, the height component is the function sin(x) (pronounced “sine of x”), and the width component is the function cos(x) (pronounced “cosine of x”). Using zero as a value for “x” in both of these equations will yield our original width and height of one and zero, respectively.
We’ll come back to this subject later once we explore their similarities with the hyperbolic functions. Until then, we also have to review the all-important exponential function.
The exponential function is one of the single most essential functions in all of mathematics. We won’t describe its full capabilities and applications here, but for now, understand that it adequately describes the growth of just about anything in the natural world: plant growth, radioactive decay and numerous other phenomena are all accurately quantified using the exponential function.
The exponential function is represented by y=ex, where e=2.71828…. On a graph, this forms a hyperbola that grows at an exponential rate. e, also known as “Euler’s number,” just-so-happens to be a near-universal ratio of growth.
This function is (hopefully) interesting and all, but how does it relate to hyperbolic functions?
Image Source : Pixabay
Hyperbolic functions describe the hyperbola of the exponential function in a similar way that trigonometric functions describe the unit circle. The names of the basic “height” and “width” functions are remarkably similar to those from trigonometry!
Where trigonometric functions include sin x and cos(x), hyperbolic functions include sinh(x) (pronounced “cinch of x”) and koshx (pronounced “kosh of x” or “kosh” rhymes with “gosh”). These are defined using the exponential function such that:
sinh x =ex-e-x2 and x =ex+e-x2
Note the subtle difference between the functions: The numerator of the sinh(x) function is ex-e-x, while that of the koshx function is ex+e-x.
The names and functions of the hyperbolic functions aren’t their only similarities to trigonometry; other hyperbolic functions and identities are, in fact, remarkably similar to their counterparts in trigonometry.
One example of this similarity is the “double angle” identity, which in trigonometry is sin 2x =2sin x cos x . What happens if we work backward and replace sin x and cos(x) with their hyperbolic cousins? Something interesting happens:
2sinh x x =2ex-e-x2ex+e-x2=e2x-e-2×2=sinh 2x
The identity is the same as it is in trigonometry! Other trigonometric parallels are also possible but are somewhat beyond the scope of this guide.
For now, we’ve defined the two most basic hyperbolic functions— sinh(x) and koshx. At this point, you’ve probably determined that finding a hyperbolic derivative involves finding the derivative of one of these (or one of the other) functions.
Before actually calculating hyperbolic derivatives, however, we’ll have a quick review on regular derivatives. This review will be especially useful, as the exponential function has some very unique properties when it comes to derivation.
Again, a derivative of a function is its rate of change with a variable. Returning to the car analogy, we might be able to describe a car’s distance traveled as a function of time. In other words, we could have a function that says our car moves a distance of “x” over time “t” such that x=50t.
If we assign “x” to have its unit in meters and “t” to have its unit in seconds, we could say that our car will cover a distance of 50 meters in one second (which, by the way, is very fast—about 112 miles per hour!).
We might be curious to know, then, how fast the car is going. Intuition tells us that if the car travels 50 meters in one second, it must have a speed of 50 meters per second. Just by going through this mental exercise, you’ve calculated a derivative!
The exact theory of derivation is well beyond the scope of this guide. For now, remember that a derivative of a function describes that functions rate of change.
When solving a derivative algebraically, one of the most powerful rules to remember is the “power rule.” Here, “power” refers to whatever power a function’s variable is raised to. For example, the function y=x2 has the variable “x” raised to a power of two.
What would be the derivative of such a function? Or, in other words, what would be its rate of change as we change the value of “x?” The power rule offers a convenient solution, stating that the derivative of xn=n*xn-1. So, in the case of y=x2, dydx=2x (here, dydx means “the derivative of function ‘y’ by variable ‘x’”).
What if we try that on our car’s position equation from earlier? A quick calculation shows that x’=1*50t1-1=50, which is exactly what our intuition told us.
This rule works for most algebraic equations, but some functions have special derivation rules. One of these is the exponential function from earlier, whose derivative is – wait for it – itself! Formally stated, this relationship takes the form dydxenx=nenx. Take special care to remember any constant in front of the variable “x!”
As you might be able to imagine, this relationship is especially helpful for calculating hyperbolic derivatives—which we can now finally do.
Image Source : Pixabay
Now that we have a solid background, we can move forward to deriving our essential hyperbolic functions. With the knowledge from the previous sections, the derivations should be reasonably straightforward.
For calculating the derivative of sinh x , we derivate its value of ex-e-x2. Using the rules described in the previous section, this yields an exciting result: ex+e-x2. This value looks a lot like x because it is! Similarly, deriving x will produce the value of sinh x .
It appears that the derivatives of the two essential hyperbolic functions sinh x and x are, in fact, each other. Remembering the parallels between hyperbolic and trigonometric identities, one can easily derive hyperbolic functions such as tanh x , where tanh x =sinh(x)kosh(x) just as tan x =sin(x)cos(x). Can you find its derivative?
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]]>The post What Is the Isosceles Triangle Theorem? appeared first on Calculus Help.
]]>Before we cover the Isosceles Triangle Theorem, we’ll discuss how we have used triangles over time in architecture, art, and design. Then we’ll talk about the history of isosceles triangles, the different types of triangles, and the different parts of isosceles triangles.
And lastly, after discussing the theorem, we’ll go over some useful formulas for calculating various parts of isosceles triangles.
While rectangles are more prevalent in architecture because they are easy to stack and organize, triangles provide more strength. With modern technology, triangles are easier to incorporate into building designs and are becoming more prevalent as a result.
In 1989, Japanese architects decided that a triangular building design would be necessary if they were to construct a 500-story building in Tokyo. The triangular shape could withstand earthquake forces, unlike a rectangular or square design.
You can also see triangular building designs in Norway, the Flatiron Building in New York, public buildings and colleges, and modern home designs.
As far as isosceles triangles, you see them in architecture, from ancient to modern. Ancient Egyptians used them to create pyramids. Ancient Greeks used obtuse isosceles triangles as the shapes of gables and pediments. In the Middle Ages, architects used what is called the Egyptian isosceles triangle, or an acute isosceles triangle.
You can also see isosceles triangles in the work of artists and designers going back to the Neolithic era. They are visible on flags, heraldry, and in religious symbols.
The study of triangles is almost as old as civilization. Ancient Egyptians studied them as did Babylonians. You can see how ancient Egyptians used triangles to construct pyramids.
It wasn’t until about 300 BCE that the Greek mathematician, Euclid, gave triangles with two equal sides a name. He combined the Greek words “isos” (equal) and “skelos” (legs) to define them as triangles with exactly two sides.
Today, mathematicians define isosceles triangles as having at least two equal sides. This is a subtle but important difference because it means that equilateral triangles are also considered to be isosceles triangles. More on that below.
In the world of geometry, there are many types of triangles besides isosceles:
Isosceles triangles can also be acute, obtuse, or right, depending on their angle measurements. Equilateral triangles can be a type of isosceles triangle. However, note that not all isosceles triangles are equilateral. Moreover, an isosceles triangle can never be a scalene triangle.
There are a few particular types of isosceles triangles worth noting, such as the isosceles right triangle, or a 45-45-90 triangle. There is also the Calabi triangle, an obtuse isosceles triangle in which there are three different placements for the largest square.
And last but not least, there is also the golden triangle, which is an isosceles triangle where the duplicated leg is in the golden ratio to the distinct side. The golden ratio is defined as a ratio of two numbers in which the ratio of the sum to the bigger number is the same as the ratio of the larger number to the smaller.
No matter how you define isosceles triangles, they are all made up of two legs and a base. If it’s an equilateral triangle, all sides can be considered the base because all sides are equal.
And, there are two equal angles opposite the equal sides. Each of these angles is called a base angle. The angle in between the legs is the vertex angle.
Also, the angles opposite each leg are equal and always less than 90 degrees (acute). All total, the angles should add up to 180 degrees.
Finally, it’s time to discuss the Isosceles Triangle Theorem. The Isosceles Triangle Theorem states: In a triangle, angles the opposite to the equal sides are equal.
So, how do we go about proving it true? It’s pretty simple. First, we’re going to need to label the different parts of an isosceles triangle.
Let’s give the points of the isosceles triangle the labels A, B, and D (counterclockwise from the top). We also need to draw a line from the center of the base (BD) to the angle (A) on the other side. Note that the center of the base is termed midpoint, and angles on the inside of the triangle are called interior angles.
Where that line intersects the side is labeled C. The line creates two triangles, ABC and ACD.
So, Point C is on the base BD, creating line segment AC. Here’s what we have so far:
We have what is called the Side Side Postulate because all of the sides of ABC (three total) are congruent with ACD. If triangle ABC and triangle ACD have congruence, then their matching parts are congruent. This also proves that the B angle is congruent with the D angle.
As with most mathematical theorems, there is a reverse of the Isosceles Triangle Theorem (usually referred to as the converse). It states, “if two angles of a triangle are congruent, the sides opposite to these angles are congruent.” Let’s work through it.
First, we’ll need another isosceles triangle, EFH. The EFH angle is congruent with the EHF angle. We need to prove that EF is congruent with EH. To do that, draw a line from FEH (E is the apex angle) to the base FH. Label this point on the base as G.
By doing this, we have made two right triangles, EFG and EGH. Because we have an angle bisector with the line segment EG, FEG is congruent with HEG. So what is the result?
We now have what’s known as the Angle Angle Side Theorem, or AAS Theorem, which states that two triangles are equal if two sides and the angle between them are equal.
Let’s take a look. We know that EFG is congruent with EHF. FEG is congruent with HEG. And EG is congruent with EG. That gives us two angles and a side, which is the AAS theorem.
When the triangles are proven to be congruent, the parts of the triangles are also congruent making EF congruent with EH. By working through everything above, we have proven true the converse (opposite) of the Isosceles Triangle Theorem.
In addition to understanding the Isosceles Triangle Theorem, you should also be familiar with a few basic equations for isosceles triangles.
You can use the Pythagorean theorem to find the height of any triangle. First, label the two equal sides as a, and the base as b. The height (h) equals the square root of b^{2} – 1/4 a^{2}. You can also divide the square root of 4a^{2} – b^{2}in half and get the same result.
To measure the length of the outside of a triangle, add the length of each side together. But, since isosceles triangles have two equal sides, you can make the process easier with this formula: p = 2a + b.
Area is defined as the total of unit squares you can fit inside any given shape. For an isosceles triangle, divide the total of the base (b) x height (h)by 2.
If you don’t know the height, use the formula listed above to calculate it.
You can find the altitude of the isosceles triangle given the base (B) and the leg (L) by taking the square root of L^{2} – (B/2)^{2}.
To find the base of an isosceles triangle when you know the altitude (A) and leg (L), it is 2 x the square root of L^{2} – A^{2}.
When given the base (B) and altitude (A), the leg is the square root of A^{2} + (B/2)^{2}.
When you know one interior angle of an isosceles triangle, it’s possible to find the other two. Let’s say that the angle at the apex is 40 degrees. Because angles must add up to 180 degrees, the two base angles need to add up to 140. Given they must be congruent angles, each of them must be 70 degrees.
In this article, we have covered the history of isosceles triangles, the different types of triangles, useful formulas, and various applications of isosceles triangles.
We also discussed the Isosceles Triangle Theorem to help you mathematically prove congruent isosceles triangles. For a little something extra, we also covered the converse of the Isosceles Triangle Theorem. You should be well prepared when it comes time to test your knowledge of isosceles triangles.
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]]>The post Indeterminate Forms in Calculus: What are They? appeared first on Calculus Help.
]]>Image from Pixabay
Indeterminate forms in calculus begin with algebraic functions that utilize a limit for the independent variable to find a solution. The indeterminate aspect of the form results when different rules vie to apply to the solution.
Since there is no clear way to determine which rule of calculus will govern the answer, the function then becomes an intermediate form. There are many indeterminate forms, but there is only a handful that commonly occurs. For instance, 0/0 and 0 to the power of 0 are examples of more common indeterminate forms.
Replacing algebraic combinations within a function with its limits, assuming an independent variable sometimes results in an answer such as 0/0. In this case, there are several competing rules vying for an opportunity to define this solution.
Indeterminate forms, officially coined by a student of the famed French mathematician Augustin Cauchy, have been around for as long as calculus. However, they have only been studied in the last 150 years or so.
Substituting a limit that results in a zero, infinity, negative infinity, or any combination of these may result in an indeterminate form. When both functions approach the given limit that results in the indeterminate form, there is not enough information to determine what the behavior of the function is at that point.
Some indeterminate forms can be solved by factoring through elimination or using L’Hopital’s rule. The functions resulting in 0/0 and infinity over negative infinity can achieve a solution through various means.
An indeterminate form is a limit that is still easy to solve. It only means that in its current form as a limit put into a function, it presents too many unknowable characteristics to form an appropriate answer properly. You can’t just solve for the quotient.
When solving for a limit, we are looking at two functions so that they make a ratio. Ratios are the most common, but not the only, way to discern an indeterminate function. Several geometric functions are also indeterminate forms, but not ratios.
Finally, while limits resulting in zero, infinity, or negative infinity are often indeterminate forms, this is not always true. Infinity, negative or positive, over zero will always result in divergence. As well, one over zero has infinite solutions and is therefore not indeterminate. We can determine a universal solution, but an indeterminate answer is one that needs more information.
Indeterminate forms hover over the calculus no matter where you turn. When dealing with ratios such as 1/0, 0/0, or infinity in any form, you will most likely need to use a further theorem, such as L’Hopital’s, to solve for a limit. Knowing that indeterminate forms are sometimes solvable can elicit clarity in a function that previously was not available.
Delving further into the particulars of indeterminate forms allows us to utilize a variety of methods to determine the characteristics of a function at a particular limit. Not enough information determines an indeterminate form. However, sometimes, we have too much information and need to whittle down a solution to one technique.
Take the fraction 0/0. An elementary view of this fraction would tell someone that it equals one because a numerator that equals a denominator equals one. Or more practically, we could think of the fraction as just zero. Further, a zero in the denominator could indicate infinity or does not exist. Therefore, there are options for classifying this ratio, but no clear winner stands out.
So why are indeterminate forms necessary in practical, real-life situations? The answer is that they aren’t, at least not by themselves. However, their mere existence provides the ground for individuals to find ways to make indeterminate forms solvable.
Many areas of physics and math will use a denominator of zero as a potential starting point in a variety of situations and formulas. Laws involving physics, heat, and quantum mechanics will use indeterminate forms at some point. Understanding these forms allows you to solve them using an appropriate method, whether that is L’Hopital’s method or another.
From an even more practical standpoint, there could be very general problems involving motion, velocity, and time that require the use of indeterminate forms. Questions surrounding the speed and time of a moving object will have to start at zero. Determining an infinite limit of two of these objects would result in an indeterminate form.
Again, it is critical to note the reason for classifying an indeterminate form as “indeterminate.” It is to differentiate it from other ratios that are zero or does not exist. To alleviate confusion, possibly substituting the word “temporary” for indeterminate would clear up some of the misconceptions surrounding the use of these forms.
Understanding these forms as a transient is a better way to think of them. An indeterminate form, therefore, is just a vehicle for further computation that is up to the discretion of the user.
In the real-world, an indeterminate form would take the form of algebraic expression with a defined limit. Let’s say the numerator was x-4, and the denominator was 2x-8. With a limit of 8, what is the value of the ratio? The ratio becomes 0/0.
Once again, 0/0 is an indeterminate form. Solving it further requires some factoring. If we cancel out the variable, we get a ratio of ⅔. Using simple algebra, we can get away from the indeterminate form.
When simple algebra is unavailable for use in solving an indeterminate form, then L’Hopital’s rule becomes necessary. A simple search of the internet results in some handy lists of indeterminate forms.
Typically these lists are arrayed into three columns. The first column lists the indeterminate form. The second shows how to use L’Hopital’s rule to get a ratio to either 0/0 or infinity/infinity. Once the ratio becomes either of those, then they are solved either algebraically or using derivatives.
While L’Hopital’s rule is not precisely real-world applicable on its own, it is also a vehicle for further calculation much like indeterminate forms. Without the rule and the existence of indeterminate forms, calculations for meaningful and applicable formulas in other areas would not be possible.
How do we know when an indeterminate form needs the use of L’Hopital’s rule or needs to use algebra?
When using indeterminate forms, be sure you have a proper understanding of what a derivative is and how to use them. The first step in finding the derivative of a polynomial is to bring each exponent down by one. Multiply the initial exponent by the coefficient. Multiplying allows you to find the derivatives and use L’Hopital’s formula.
It is critical to forming a thorough understanding of L’Hopital’s formula if you will be dealing with indeterminate forms. Instances where you could get end up with zero minus infinity, or infinity minus infinity all call for the use of this rule. The use of derivatives allows you to transform this indeterminate form into 0/0 or infinity over infinity. From there, you can solve using algebra.
There are instances where L’Hopital’s rule will not work with an indeterminate form. If you have zero over infinity, finding the derivative of the polynomials will only lead to the wrong answer. Using the inverse of the infinite polynomial will then change the ratio into an L’Hopital-friendly 0/0 ratio.
Instances, where a function equals zero to the zero power, requires the use of natural logarithms. Subtracting to infinities calls for using the laws of trigonometry and making calculations using cos, sin, and tan. Any indefinite forms that you find in the course of your calculus journey have a method for solving.
While an indeterminate form may not change your life, it can provide the means for further understanding of broader calculus concepts and rules. As well, indeterminate forms are primarily made up of infinity, zero, and one, which is the primary values often dealt with in calculus. Understanding their indeterminate forms is crucial.
Many subject areas will use formulas and calculations involving indeterminate forms. A thorough understanding of how to solve them, and the parts that make up their solutions such as derivatives and trigonometric functions, are fundamental.
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